Approximations in Chemical Equilibrium (add a weak acid HA into pure water)

AI Thread Summary
The discussion focuses on the equilibrium dynamics of a weak acid (HA) added to pure water, considering the autoionization of water and the associated equilibrium constants (K_a and K_w). The initial concentrations of hydronium (H3O+) and hydroxide (OH−) ions due to water's autoionization are acknowledged, leading to a set of equilibrium expressions. The presence of H3O+ affects the dissociation of HA, suggesting that the reverse association process occurs more rapidly, thus limiting the concentration of H3O+ and resulting in a lower concentration of OH− at equilibrium compared to initial conditions.The conversation also critiques the common practice of neglecting initial concentrations of H3O+ and OH− in equilibrium calculations, arguing for their inclusion to enhance accuracy. The derived cubic equations for the concentrations at equilibrium are discussed, with a focus on simplifying these equations under certain assumptions, such as the "5% rule," which allows for approximations that reduce the complexity of calculations.
SilverSoldier
Messages
26
Reaction score
3
Suppose we add a weak acid HA into pure water, so that upon addition its initial concentration is c. The following equilibria should establish in the system. $$\text{HA}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{A}^-$$ $$2\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{OH}^−$$ Let ##K_a## and ##K_w## respectively be the equilibrium constants for these processes. We can express the composition of the system before and after equilibrium as follows. Let ##a## be the concentration of ##\text{H}_3\text{O}^+## and ##\text{OH}^−## already present in the system due to the autoionization of water initially; i.e., ##a=\sqrt{K_w}##.

##\text{HA}##​
##\text{H}_3\text{O}^+##​
##\text{A}^-##​
Initial concentration
##c##​
##a##​
-​
Change in concentration
##-p##​
##+p##​
##+p##​
Equilibrium concentration
##c-p##​
##a+p+q##​
##p##​

##\text{H}_2\text{O}##​
##\text{H}_3\text{O}^+##​
##\text{OH}^−##​
Initial concentration
-​
##a##​
##a##​
Change in concentration
-​
##+q##​
##+q##​
Equilibrium concentration
-​
##a+p+q##​
##a+q##​

Thus, we can obtain the following expressions at equilibrium. $$K_w=\left(a+p+q\right)\left(a+q\right)$$ $$K_a=\dfrac{\left(a+p+q\right)\left(p\right)}{c-p}$$ Taking ##r=a+q##, we obtain the following cubic equations in ##p## and ##r##. $$p^3K_a+p^2\left(K_a^2−K_w−cK_a\right)−2pcK_a^2+c^2K_a^2=0$$ $$r^3K_a+r^2\left(cK_a+K_w\right)−rK_aK_w−K_w^2=0$$ Now, as the acid begins to dissociate, because there is ##\text{H}_3\text{O}^+## already present in the system, the reverse "association" process should start off faster, so it should take lesser time for its rate to increase up to a significant value bringing the system to equilibrium, allowing time only for fewer molecules of the acid to dissociate.

At the same time, because ##\text{H}_3\text{O}^+## is being "removed" by the reverse autoionization process as it forms, the system should be trying to keep the ##\text{H}_3\text{O}^+## concentration below whatever value it would otherwise have risen to in the absence of autoionization.

The fact that the reverse autoionization process is triggered as the reaction proceeds should mean that ##\text{OH}^-## ions in the system should be being consumed, so by the time equilibrium is attained, its concentration must be lesser than the initial value. This means that ##r## in the equation above must be less than ##a##, so if ##a## can be considered small, then ##r^3## must be even smaller, and the effect of the ##r^3## term in the equation should be negligible. It should therefore be possible the reduce it to $$r^2\left(cK_a+K_w\right)−rK_aK_w−K_w^2=0$$ The following is a plot I made of the above equations with the ##x## axis representing ##c## and ##y## axis representing ##r##. The ##r^3## term is neglected in the orange curve, and not neglected in the black curve (##K_a=5## and ##K_w=0.5## here).

geogebra-export.png

The orange curve greatly deviates from the black curve for small concentrations. Is this possible, because ##r## must always be less than ##a## for all concentrations? Have I made an error somewhere? Is it wrong to neglect the ##r^3## term?

By the way, I have not seen calculations being made taking the ##\text{H}_3\text{O}^+## and ##\text{OH}^-## concentrations already present in the system initially into consideration. Where the autoionization of water is not neglected it is always assumed that there are no ##\text{H}_3\text{O}^+## and ##\text{OH}^-## ions present in the system before equilibrium. Why is this? Is there anything wrong with saying it is already present?
 
Last edited:
Chemistry news on Phys.org
In general when systematically finding equilibrium nobody cares about initial concentrations of anything. Total (analytical) concentrations and mass balances are all that matters (apart from Ka/Kb and charge balance).

For nitpickers one of the mass balances should be that of water itself, but it makes the calculations a bit more difficult, as it forces the solver into mass balances of elements, not of substances. The most general equilibirum programs do that, those specialized for pH calculations don't, as it is enough to add Kw to make sure autoionization of water is part of the system.

While you approach - starting with ICE tables - should in principle produce the same results, I have never seen it used that way. The general approach is actually simpler and easier to use.
 
  • Informative
Likes BillTre and berkeman
Borek said:
In general when systematically finding equilibrium nobody cares about initial concentrations of anything. Total (analytical) concentrations and mass balances are all that matters (apart from Ka/Kb and charge balance).

For nitpickers one of the mass balances should be that of water itself, but it makes the calculations a bit more difficult, as it forces the solver into mass balances of elements, not of substances. The most general equilibirum programs do that, those specialized for pH calculations don't, as it is enough to add Kw to make sure autoionization of water is part of the system.

While you approach - starting with ICE tables - should in principle produce the same results, I have never seen it used that way. The general approach is actually simpler and easier to use.
How are we able to simplify the above cubics to simpler equations that can be solved? How do we, for example, simplify the cubic in ##p## to obtain the usual result that ##p=\sqrt{cK_a}##?
 
Look here for a review of methods used: https://www.chembuddy.com/?left=pH-calculation&right=toc

Broadly speaking the main idea is the "5% rule" - if in a sum x+y one component (say y) is less than 5% of the other we try to ignore it and assume x+y≈x. That typically decreases the degree of the resulting polynomial.
 
It seems like a simple enough question: what is the solubility of epsom salt in water at 20°C? A graph or table showing how it varies with temperature would be a bonus. But upon searching the internet I have been unable to determine this with confidence. Wikipedia gives the value of 113g/100ml. But other sources disagree and I can't find a definitive source for the information. I even asked chatgpt but it couldn't be sure either. I thought, naively, that this would be easy to look up without...
I was introduced to the Octet Rule recently and make me wonder, why does 8 valence electrons or a full p orbital always make an element inert? What is so special with a full p orbital? Like take Calcium for an example, its outer orbital is filled but its only the s orbital thats filled so its still reactive not so much as the Alkaline metals but still pretty reactive. Can someone explain it to me? Thanks!!
Back
Top